Epsilon delta proof min http://www.milefoot.com/math/calculus/limits/DeltaEpsilonProofs03.htm
I've been studying these épsilon delta proofs. In the non-linear case, he gets:
$$\delta=\min\left\{5-\sqrt{25-\dfrac{\epsilon}{3}},-5+\sqrt{25+\dfrac{\epsilon}{3}}\right\}$$
Well, I know that these $\delta$ are not equal the opposite of the other, but it has shown that $x$ must be within the range covered by these two deltas. Well, I already have bounded the $x-a$ (in this case, $x-5$) in therms of $\epsilon$, so it should work that for any given $\epsilon$, i could get only the  $-5+\sqrt{25+\dfrac{\epsilon}{3}}$. Why I have to get the minimum?
 A: It is because $\delta$ has to be acceptable in the worst case.  Say we are proving $\lim_{x \to 0} f(x)=L$ and for (the given) $\epsilon$ we are within $\epsilon$ over the interval $\delta \in (-1,0.1)$  The definition of limit is symmetric:  it says whenever $x$ is within $\delta$ of $0$, then $|f(x)-L|\lt \epsilon$ so we have to shrink the interval to make it symmetric, so our answer should be within $\delta \in (-0.1,0.1)$.  This sounds restrictive, but it is not.  One can prove that symmetric limits leads to the same thing as asymmetric limits and every interval includes a symmetric interval.
A: He takes the minimum so that he can use the following two inequalities.
$$
\delta \leq 5-\sqrt{25-\dfrac{\epsilon}{3}} \\
\delta \leq -5+\sqrt{25+\dfrac{\epsilon}{3}}
$$
The first inequality above can be written as
$$
-5 + \sqrt{25-\dfrac{\epsilon}{3}} \leq -\delta.
$$
So the proof uses these inequalities to get
$$
-5 + \sqrt{25-\dfrac{\epsilon}{3}} \leq -\delta < x - 5 < \delta \leq -5+\sqrt{25+\dfrac{\epsilon}{3}}
$$
which is the inequality he intended to get (as you can see when he worked backwards). Hopefully that clears up why he needed to take $\delta$ as the minimum of the two quantities.
